Let $f:X\rightarrow X$ be a contractive mapping of a complete metric space satisfying $$d(f(x),f(y))\leq\alpha(d(x,y))d(x,y)$$ where $\alpha:\mathbf{R}^+\rightarrow [0,1)$, and $\alpha(t_n)\rightarrow 1$ implies $t_n\rightarrow 0$
Does $f$ have a fixed point?
Thank you.
It seems that the answer is yes.
We may replace $\alpha(d)$ by $\beta(d)=\sup_{t\geq d}\alpha(t)$ (surely, $\beta(q)$ is the limit of some sequence $\alpha(t_n)$ with $t_n\geq d$, so $\beta(d)<1$). Thus we obtain monotone $\alpha(d)$.
As usual, start with any $x_0\in X$ and define $x_{n+1}=f(x_n)$, $d_{n+1}=d(x_n,x_{n+1})\leq \alpha(d_n)d_n$. The monotone sequence $d_n$ converges to some $D$; if $D>0$ then we have $D\leq \alpha(D)D$ which is absurd. Hence $D=0$.
Now denote $r_n=\sup_{k>n}d(x_n,x_k)$. We want to show that $r_n\to 0$ as $n\to 0$ (in particular, this will yield that all the $r_n$ are finite). Assume, to the contrary, that there exists some $\mu>0$ such that $r_n>\mu$ infinitely often.
Set $\nu=\alpha(\mu)$. There exists an $n$ with $r_n>\mu$ such that $d_n<\frac{(1-\nu)\mu}{2\nu}$. Choose $k>n$ such that $d(x_n,x_k)>\mu$ (then $d(x_{n-1},x_{k-1})>\mu$ as well). Then we have $$ d(x_n,x_k)\leq \alpha(d(x_{n-1},x_{k-1}))d(x_{n-1},x_{k-1}) \leq \nu(d_n+d(x_n,x_k)+d_k) \leq \nu(d(x_n,x_k)+2d_n), $$ or $(1-\nu)\mu<(1-\nu)d(x_n,x_k)\leq 2\nu d_n$. This contradicts the choice of $n$.
Thus $r_n$ indeed tend to $0$; therefore, our sequence $(x_n)$ is convergent to the common point of the balls $B_{r_n}(x_n)$. Surely, this limit is the sought fixed point.
The answer is yes, if $\alpha$ is increasing. Take any $x_0$ and define $x_{n+1}=f(x_n)$. Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$ Indeed, we have $$|x_{n+1}-x_n|\leq\alpha(|x_{n}-x_{n-1}|)|x_{n}-x_{n-1}|.$$ In particular, $|x_{n+1}-x_n|\leq|x_n-x_{n-1}|$, from which follows by induction that $|x_{n+1}-x_n|\leq|x_1-x_0|.$ So we can continue the previous inequality: $$|x_{n+1}-x_n|\leq k|x_{n}-x_{n-1}|,$$ and again by induction $$|x_{n+1}-x_n|\leq k^n|x_1-x_0|,$$ Therefore the series $$\sum|x_{n+1}-x_n|$$ is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.
(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)
Fix $x\in X$ and let $x_n=f^n(x)$, $n=1,2, \ldots$ yet again we break the argument into two steps.
Step 1: $\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=0$
Since $f$ is contractive the sequence $\{d(x_n,x_{n+1})\}$ is monotone decreasing and bounded below, so $$\lim\limits _{n\rightarrow\infty}d(x_n,x_{n+1})=r\geq0$$ Assume that $r>0$. Then by the contractive condition:
$$\frac{d(x_{n+1},x_{n+2})}{d(x_n,x_{n+1})}\leq \alpha(d(x_n,x_{n+1}))\,\,\,,n=1,2,3, \ldots$$
Then $n\rightarrow\infty\implies 1\leq\lim\limits _{n\rightarrow\infty}\alpha(d(x_n,x_{n+1}))\implies r=0$, contradiction.
Step 2: $\{x_n\}$ is a Cauchy sequence:
Assume $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)>0$.
$$d(x_n,x_m)\leq d(x_n,x_{n+1})+d(x_{n+1},x_{m+1})+d(x_{m+1},x_m)$$ So by the contractive condition : $$d(x_n,x_m)\leq {(1-\alpha d(x_n,x_{m}))}^{-1}[d(x_{n},x_{n+1})+d(x_{m+1},x_m)]$$ Under the assumption $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)>0$,step 1 implies $\lim\limits_{n,m\rightarrow\infty}\sup {(1-\alpha (d(x_n,x_m)))}^{-1}=+\infty$
From which $\lim\limits_{n,m\rightarrow\infty}\sup \alpha (d(x_n,x_m))=1$, that implies $\lim\limits_{n,m\rightarrow\infty}\sup d(x_n,x_m)=0$, again a contradiction.
